Question

How do you divide $\left( 9{{x}^{3}}+5 \right)\div \left( 2x-3 \right)$ using long division?

Answer 1

Hint: We are given polynomials $\left( 9{{x}^{3}}+5 \right)$ and $\left( 2x-3 \right)$ and we are asked to divide $\left( 9{{x}^{3}}+5 \right)$ by $\left( 2x-3 \right)$ using the long division method. To solve this question, we will learn about how the terms should be defined in long division and then we will perform these steps on our polynomial $\dfrac{9{{x}^{3}}+5}{2x-3}$ and we will learn at which step we should terminate the long division.

Complete step by step answer:
We are given the polynomials $\left( 9{{x}^{3}}+5 \right)$ and $\left( 2x-3 \right)$ and we have to divide $\left( 9{{x}^{3}}+5 \right)$ by $\left( 2x-3 \right)$. Here, $\left( 9{{x}^{3}}+5 \right)$ is a polynomial of degree 3 and $\left( 2x-3 \right)$ is a linear polynomial. We know that the difference of their powers is 2, so in the quotient we will get a 2 degree polynomial. In order to divide the polynomial, we should see whether the polynomials in the numerator and denominator have been arranged in decreasing order of their power or not.
For example, let us consider $\dfrac{1+7{{x}^{2}}+{{x}^{3}}}{{{x}^{2}}+2x}$. Here we can see that the numerator is not well defined, so we will arrange it in a descending order based on their power. So, we will get the numerator after arranging it as ${{x}^{3}}+7{{x}^{2}}+1$.
Now, let us consider our question, so we have $\dfrac{9{{x}^{3}}+5}{2x-3}$. In our polynomial both the numerator and the denominator are in the correct order. Now, we will learn how to perform the long division of polynomials with the help of the following steps.
Step 1: Divide the first term of the numerator by the first term of the denominator and then put that in the quotient.
Step 2: Multiply the denominator by that value and then put that below the numerator.
Step 3: Now, subtract them to create the next polynomial.
Step 4: Repeat these steps again with the new polynomial.
Now, we have $\dfrac{9{{x}^{3}}+5}{2x-3}$. The first term of the numerator is $9{{x}^{3}}$ and that of the denominator is 2x. So, on dividing them, we will get $\dfrac{9{{x}^{3}}}{2x}=\dfrac{9{{x}^{2}}}{2}$. So, we will multiply the denominator by it and put it below the numerator and subtract.
\[2x-3\overline{\left){\begin{align}
  & 9{{x}^{3}}+5 \\
 & 9{{x}^{3}}-\dfrac{27{{x}^{2}}}{2} \\
 & \overline{\dfrac{27{{x}^{2}}}{2}+5} \\
\end{align}}\right.}\left( \dfrac{9{{x}^{2}}}{2} \right.\]
Here, the higher power is in \[\dfrac{27{{x}^{2}}}{2}\] and in 2x-3, we have 2x. So, we will divide \[\dfrac{27{{x}^{2}}}{2}\] by 2x. So, we get, \[\dfrac{27{{x}^{2}}}{2\left( 2x \right)}=\dfrac{27x}{4}\]. So, we multiply (2x-3) by \[\dfrac{27x}{4}\] and subtract that from the numerator, so we get,
\[2x-3\overline{\left){\begin{align}
  & 9{{x}^{3}}+5 \\
 & 9{{x}^{3}}-\dfrac{27{{x}^{2}}}{2} \\
 & \overline{\dfrac{27{{x}^{2}}}{2}+5} \\
 & \dfrac{27{{x}^{2}}}{2}-\dfrac{81x}{4} \\
 & \overline{\text{ }\dfrac{81x}{4}+5} \\
\end{align}}\right.}\left( \dfrac{9{{x}^{2}}}{2}+\dfrac{27x}{4} \right.\]
Now, we have the highest power is \[\dfrac{81x}{4}\] in the numerator while in 2x-3, we have 2x. So, dividing \[\dfrac{81x}{4}\] by 2x, we get \[\dfrac{81x}{4\left( 2x \right)}=\dfrac{81}{8}\]. So, now multiplying \[\dfrac{81}{8}\] by 2x-3 and then subtracting, we get,
\[2x-3\overline{\left){\begin{align}
  & 9{{x}^{3}}+5 \\
 & 9{{x}^{3}}-\dfrac{27{{x}^{2}}}{2} \\
 & \overline{\dfrac{27{{x}^{2}}}{2}+5} \\
 & \dfrac{27{{x}^{2}}}{2}-\dfrac{81x}{4} \\
 & \overline{\begin{align}
  & \text{ }\dfrac{81x}{4}+5 \\
 & \text{ }\dfrac{81x}{4}-\dfrac{283}{8} \\
 & \overline{\text{ }\dfrac{283}{8}} \\
\end{align}} \\
\end{align}}\right.}\left( \dfrac{9{{x}^{2}}}{2} \right.+\dfrac{27x}{4}+\dfrac{81}{8}\]
Here, the degree of the numerator is less than that of that denominator, so we cannot divide them further. So, we get our solution as \[\dfrac{9{{x}^{2}}}{2}+\dfrac{27x}{4}+\dfrac{81}{8}\] and the remainder as $\dfrac{283}{8}$.

Note:
One has to take care while subtracting two values, as the sign of the terms will get changed, so if one uses the wrong sign at some point in the division, then the solution will become wrong. Also, one can divide higher power by lower power terms, and if we are left with the denominator with higher degree than the numerator, then division will stop.